The Foundations of Chemistry

Phosphoric acid is a typical weakpolyprotic acid. Let us now describe solutions of
sulfuric acid, a very strongpolyprotic acid.

EXAMPLE 18-17 Solutions of Strong Polyprotic Acid

Calculate concentrations of all species present in 0.10 MH 2 SO 4. Ka21.2 10 ^2.

Plan

Because the first ionization step of H 2 SO 4 is complete, we read the concentrations for the first
step from the balanced equation. The second ionization step is notcomplete, and so we write
the ionization equation, the Ka2expression, and the algebraic representations of equilibrium
concentrations. Then we substitute into Ka2for H 2 SO 4.

Solution

As we pointed out, the first ionization step of H 2 SO 4 is complete.

100%

H 2 SO 4 
H 2 O8888n H 3 O

HSO 4 

0.10 M ::::::F 0.10 M 0.10 M

The second ionization step is not complete, however.

HSO 4 
H 2 O 34 H 3 O

SO 42  and Ka21.2 10 ^2

Let x[HSO 4 ] that ionizes. [H 3 O
] is the sum of the concentrations produced in the first
and second steps. So we represent the equilibrium concentrations as

HSO 4  
H 2 O 34 H 3 O
 
SO 42 

(0.10x) M (0.10
x) Mx M

from 1st step from 2nd step

Substitution into the ionization constant expression for Ka2gives

Ka21.2 10 ^2 

Clearly, xcannot be disregarded because Ka2is too large. This equation must be solved by the
quadratic formula, which gives x0.010 and x0.12 (extraneous). So [H 3 O
]2nd
[SO 42 ]0.010 M. The concentrations of species in 0.10 MH 2 SO 4 are

[H 2 SO 4 ]
 0 M [HSO 4 ]  (0.10x) M  0.09 M [SO 42 ]0.010 M

[H 3 O
]  (0.10
x) M  0.11 M

[OH]9.1 10 ^14 M

In 0.10 MH 2 SO 4 solution, the extent of the second ionization step is 10%.

You should now work Exercise 62.

In Table 18-8 we compare 0.10 Msolutions of these two polyprotic acids. Their acidi-
ties are very different.

1.0 10 ^14



0.11

Kw



[H 3 O
]

(0.10
x)(x)



0.10x

[H 3 O
][SO 42 ]



[HSO 4 ]

[H 3 O
][SO 42 ]



[HSO 4 ]

18-5 Polyprotic Acids 775